Mat 2 - Calculus for Management and Social Science. Understanding te basics of lines in te -plane is crucial to te stud of calculus. Notes Recall tat te and -intercepts of a line are were te line meets te and aes respectivel. Also te slope of a line is te ratio of te vertical cange to te orizontal cange between an two points on te line. Sometimes people sa tat slope is te rise to run. Parallel lines ave te same slope. Perpendicular lines ave slopes tat are opposite reciprocals of eac oter. Lets look at a couple different forms for te equation of a line. Slope-Intercept Form: A nonvertical line L as an equation of te form = m + b, were m is te slope and te point (0, b) is te -intercept. For eample: = 3 2 3. 2 6 2 2 6 2 Te slope of tis line is 3 and its -intercept is (0, 3). For ever two units to te rigt, te 2 line rises tree units verticall. Point-Slope Form: A nonvertical line L as an equation of te form = m( ), were m is te slope and te point (, ) is on te line. Later wen we stud tangent lines to curves, we will see tat tis is an etremel nice form for te equation of a line.
For eample, find te equation of te line tat passes troug (, 3) and (0, ): Well, calculate slope directl: vertical cange ( 3) m = = = 2 orizontal cange 0 6 = 3. Ten we can give te equation as ( 3) = ( ), 3 + 3 = ( ). 3 Eercise: Suppose ou graduate and get a job wit Firm X. Your income, t ears after being ired, is modeled b I = 3, 500t + 0, 000. Wat is te significance of te slope and I-intercept for tis line?.2 Te slope of a curve at a point is te slope of te tangent line to te curve at te point. Te grap of f() = 2 and its tangent line at (, ): 2 6 2 2 6 2 Tis slope of a curve at a point is te rate of cange in te function as te curve passes troug tat point. We can see in te above grap tat te slope of tis curve varies greatl from point to point. At (, ) te slope is 2. At ( 2, ) te slope is. Witout deriving anting et, notice anting? Te slope of f() = 2 at (, 2 ) appears to be twice te -coordinate. In fact: Te slope of f() = 2 at te point (, 2 ) is 2.
Lets find te equation of te tangent line to f() = 2 over = 3. First of all, te slope is m = 2(3) = 6. Secondl, te point of tangenc is (3, 3 2 ) = (3, 9). Hence te equation of te tangent line is.3 9 = 6( 3). Te function tat gives te slope of f() at an arbitrar point is called te derivative of f() and is denoted b f () or df. Te process of computing te derivative is called d differentiation. As we ave alread seen, if f() = 2, ten f () = 2. Lets look at some standard functions and teir derivatives: f() f () b 0 m + b m 2 2 3 3 2 n n n Tis last rule is known as te power rule for differentiation and we postpone its proof. Wat does tis rule sa f() =? f() = = 2 = f () = 2 2 = 2. Wat about g() =? g() = = = g () = 2 = 2. Wat is te slope of g() at (2, 2 )? From above we see tat it is g (2) =. So te tangent line is 2 = ( 2). In general... Te equation of te tangent line to f() at (a, f(a)) is given b f(a) = f (a)( a).
Te limit definition of te derivative: Te difference quotient, f( + ) f(), is te slope of te secant line as can be seen in te grap above. As gets closer to zero, te secant line approaces te tangent line and te slope approaces f (). Lets return to f() = 2 and justif tat f () = 2: f( + ) f() = ( + )2 2 = 2 + 2 + 2 2 = 2 + 2 So as approaces zero, te slope of te secant lines approac f () = 2. Eercise: Justif for g() = tat g () = 2. = 2 +.. Consider te following simple function: Notice te following range values: () (2) (3) () f() = +. f() = 2 =.5 f(0) = 0.9 f(00) = 00 0.99 f(000) = 000 00.999
Notice anting? Do te range values tend to get closer to some number as we evaluate at larger and larger domain points? In fact te values are getting closer to. So we sa tat te limit of f() as approaces is. Here is te notation: lim f() =. Limits are fundamental to te stud of calculus. As we will see during our course of stud, te derivative and integral are defined in terms of limits. Te limit of a function at a domain point (or at ± ) eiter does not eist or is te value tat te function approaces as one gets close to te domain point (or ± ). Tere are, of course, rigorous definitions for te different tpes of limits tat we will stud, but we will not worr about tem. Lets see a simple eample of a limit tat does not eist. Consider te following piece-wise defined function: f() = { if 2 if > 2 Notice tat to te left of = 2 te function is constantl wile on te rigt of = 2 te function is constantl. Te function does not approac a single range value as gets close to 2, so: lim f() does not eist. 2 Limit Teorems: If lim a f() = L and lim a g() = M, ten () If k is a constant, ten lim kf() = kl. a (2) If r is positive and [f()] r is defined for a ten lim a [f()]r = L r. (3) () (5) If M 0, ten lim[f() ± g()] = L ± M. a lim[f()g()] = LM. a f() lim a g() = L M.
So, in our new language, if te derivative eists, ten f f( + ) f() (). 0 Let f() = +. Ten, using tis definition, f () 0 f( + ) f() 0 0 ++ + ( + + ) + ( + 0 + + + ( ) + + + 0 + + + ( 0 0 0 ( 0 ( = ( ( ) + + + + + ( ) + + + + + + + + ) ( + ) ( + + ) + + + ( + + + + ) ) + + + + ( + + + + ) ) + + + ( + + + + ) ) + + + ( + + + + ) ) + + 0 + ( + + + 0 + ) = ( + )(2 + ) = 2( + ) 3 2 Conversel, lets matc te following limit wit a derivative and find its value: lim 0 ( + ) 6. If te limit definition for f() = 6 is applied at =, ten f () 0 f( + ) f() ( + ) 6. 0
B te power rule f () = 6 5, so f () = 6 and tat implies ( + ) 6 lim 0 = 6. Eercise: Tr tis for lim 0 ( + )3 8..5 A function f() is continuous at = a if f(a) a f(). Tis means tat te limit eists and is equal to te value of te function. One intuitive wa to determine if a function is continuous is to ask, can tis be drawn witout lifting a pen from te page? Te function is continuous if and onl if te answer is es. Tere are tree was in wic a function can be discontinuous: Removable discontinuit: Te limit eists at = a, but it is not equal to te value of te function (wic ma or ma not be defined). Here is te grap of f() = 2. 2 3 2 2 3 Te limit of f() at = is, but f() is not defined. 2
Jump discontinuit: Te limit does not eist at = a, but as ou approac from one side or te oter tere are different limiting values. Here is te grap of g() = { 2 2 + if < 3 if > 5 0 5 3 2 2 3 Te limit does not eists at =, but approacing from te left of te function goes toward 0 wile approacing from te rigt of te function goes toward 2. Essential discontinuit: Tere is a vertical asmptote at = a. Here is te grap of 25 () = 0( ) 2. 20 5 0 5 0 2 3 Tere is a vertical asmptote at =. A function f() is continuous on an interval (a, b) if it continuous at ever point inside te interval.
A function f() is differentiable at = a if f(a + ) f(a) lim 0 eists as a real number. Tere are two was tis can go wrong for continuous functions. One is tat te limit does not eist and tere is no tangent line. Te oter is tat te tangent line is vertical. Te following are eamples of nondifferentiable functions. Te first as no tangent line at = 3, wile te second as a vertical tangent at = 3. 6 5 3 2 3 5 3 2 0 2 3 5 Teorem I : If f() is differentiable at = a ten it is continuous at = a. Note tat te converse is not true, just look at te graps above!
proof of teorem I : It is sufficient to sow tat lim(f() f(a)) = 0. a Let = a + wit 0. Ten a is equivalent to 0. Tus ( ) f(a + ) f(a) f(a + ) f(a) lim(f() f(a)) lim = f (a) 0 = 0 a 0 0 0.6 If f(), g() are differentiable and k, r are constants ten, () d d (k f()) = kf () (constant multiple rule) (2) d d (f() ± g()) = f () ± g () (sum/diff. rule) (3) d d [f()]r = r[f()] r f () (general power rule) Differentiate: Note: tis last rule is a special case of te cain rule wic we will introduce later. () + + 5 +. (2) (3 2 ) 2. (3) If () = [f()] 2 + g() ten find () given tat f() =, g() =, f () = and g () = 3. Suppose tat f() is a function and P = (a, f(a)) is a point on its grap. Tere are two important properties of te tangent line at P : Te tangent line as slope f (a). So f (a) is te rate of cange in f() at = a. Te tangent line is te line tat best approimates te grap of f() near P..7 Te second derivative of a function is te derivative of te derivative, if it eists. In our notation te second derivative is f () or d2 f d = d ( ) df. 2 d d For eample, let f() = 2 + 2. Ten f () = 2 2 3 and f () = 2 + 6.
In economics, te concept of marginalit is based on a tangent line approimation. If C() is a cost function, ten te marginal cost function is C (). Te marginal cost of producing a units, C (a), approimates C(a + ) C(a), wic is te actual additional cost of producing an etra unit. Of course tis concept etends to revenue and profit. A compan find tat te revenue R generated b spending dollars in advertising is R() = 000 + 80.02 2. Find and interpret te marginal revenue at, 000 dollars: R () = 80.0, so R (000) = 0. So te marginal revenue is 0. Tat means tat for an etra dollar of advertising, tere will be approimatel 0 dollars of etra revenue..8 Te average rate of cange in a continuous function over te interval [a, b] is measured b te slope of te secant line: f(b) f(a). b a Te derivative gives te instantaneous rate of cange. Suppose tat f() = 5. Wat is te average rate of cange of f() over te interval [, 5]? It is te slope of te secant line: f(5) f() 5 = 5 5 ( 5 ) Wat is te instantaneous rate of cange at = 3? Tat is given b te derivative: =. f () = 5 2 so ten f (3) = 5 9. Liquid water is pumped into a tank. After t minutes tere are 5t + t gallons of water in te tanks. At wat rate is te water being pumped into te tank after minutes? Let V (t) = 5t + t. Ten V (t) = 5 + 2 t and so V () = 5.25. Water is being pumped at 5.25 gallons per minute into te tank.
A particle moves in a straigt line suc tat its position after t seconds is s(t) = t 2 + 3t + 2 feet to te rigt of 0. Wat position and velocit does te particle ave after 5 seconds? Is it moving towards 0? s(5) = 25 + 5 + 2 = 2. s (t) = 2t + 3, so s (5) = 0 + 3 = 3. So te position is 2 feet to te rigt of 0, and te particle is moving 3 feet per second to te rigt, wic is awa from 0. Let C() be te cost (in dollars) of producing cars. Lets interpret C(00) = 800, 000 and C (00) =, 000: te cost of producing 00 cars is 800, 000 dollars and eac additional car costs approimatel, 000 dollars. So, for eample, producing 99 cars would cost approimatel 796, 000 dollars.